Chapter 10 —— Covariance-Stationary Vector Processes

从这一章开始我们将正式介绍向量时间序列（vector time series），Chapter 10 将聚焦于多变量动力系统理论，Chapter 11 将着重说明向量回归的估计和解释问题。

10.1 Introduction to Vector Autoregressions

$VAR(p)$ 的基本形式，并讨论协方差平稳条件下的均值表达。

${y}_{t}$ $AR(p)$ 形式为：

\begin{matrix} (10.1.1) & y_{t} = c + ϕ_{1} y_{t - 1} + ϕ_{2} y_{t - 2} + \dots + ϕ_{p} y_{t - p} + ε_{t} . \end{matrix}

其中

\begin{aligned} (10.1.2) & E (ε_{t}) & = 0 \\ (10.1.3) & E (ε_{t} ε_{τ}) & = {\begin{cases} σ^{2} & for t = τ \\ 0 & otherwise \end{cases} \end{aligned}

${\mathbf{y}}_{t}$ $n \times 1$ $VAR(p)$ 是 [10.1.1]-[10.1.3] 的向量化：

\begin{matrix} (10.1.4) & y_{t} = c + Φ_{1} y_{t - 1} + Φ_{2} y_{t - 2} + \dots + Φ_{p} y_{t - p} + ε_{t} . \end{matrix}

$\mathbf{c}$ $\left( {n \times 1}\right)$ ${\boldsymbol{\Phi }}_{j}$ $\left( {n \times n}\right)$ $j = 1,2,\ldots , p$ ${\boldsymbol{\varepsilon }}_{t}$ 的均值与协方差为

\begin{aligned} (10.1.5) & E (ε_{t}) & = 0 \\ (10.1.6) & E (ε_{t} ε_{τ}^{'}) & = {\begin{cases} Ω & for t = τ \\ 0 & otherwise, \end{cases} \end{aligned}

$\boldsymbol{\Omega }$ $\left( {n \times n}\right)$ 矩阵。

${c}_{i}$ $\mathbf{c}$ $i$ ${\phi }_{ij}^{\left( 1\right) }$ ${\boldsymbol{\Phi }}_{1}$ $i$ $j$ 列元素，则 [10.1.4] 的第 1 行为

\begin{aligned} y_{1 t} & = c_{1} + ϕ_{11}^{(1)} y_{1, t - 1} + ϕ_{12}^{(1)} y_{2, t - 1} + \dots + ϕ_{1 n}^{(1)} y_{n, t - 1} \\ + ϕ_{11}^{(2)} y_{1, t - 2} + ϕ_{12}^{(2)} y_{2, t - 2} + \dots + ϕ_{1 n}^{(2)} y_{n, t - 2} \\ (10.1.7) & + \dots + ϕ_{11}^{(p)} y_{1, t - p} + ϕ_{12}^{(p)} y_{2, t - p} + \dots + ϕ_{1 n}^{(p)} y_{n, t - p} + ε_{1 t} . \end{aligned}

$p$ $p$ 个滞后，各回归的解释变量集合相同。

$L$ 表示，可将 [10.1.4] 写为

[I_{n} - Φ_{1} L - Φ_{2} L^{2} - \dots - Φ_{p} L^{p}] y_{t} = c + ε_{t}

或

Φ (L) y_{t} = c + ε_{t},

$\boldsymbol{\Phi } \left( L\right)$ $\left( {n \times n}\right)$ $i$ $j$ 列为

Φ (L) = [δ_{i j} - ϕ_{i j}^{(1)} L^{1} - ϕ_{i j}^{(2)} L^{2} - \dots - ϕ_{i j}^{(p)} L^{p}],

${\delta }_{ij}$ $i=j$ 时为 1，否则为 0。

$E\left[{\mathbf{y}}_{t}\right]$ $E\left[{{\mathbf{y}}_{t}{\mathbf{y}}_{t - j}^{\prime }}\right]$ $t$ 变，则对 [10.1.4] 取期望得到均值

\begin{aligned} μ & = c + Φ_{1} μ + Φ_{2} μ + \dots + Φ_{p} μ, \\ μ & = {(I_{n} - Φ_{1} - Φ_{2} - \dots - Φ_{p})}^{- 1} c . \end{aligned}

将 [10.1.4] 改写为“偏离均值”的形式得到

\begin{aligned} (10.1.8) & (y_{t} - μ) & = Φ_{1} (y_{t - 1} - μ) + Φ_{2} (y_{t - 2} - μ) + \dots + Φ_{p} (y_{t - p} - μ) + ε_{t} . \end{aligned}

$VAR(p)$ $VAR(1)$

$AR(p)$ $VAR(1)$ 形式有助于分析。为此，定义

\begin{aligned} (10.1.9) & \underset{(n p \times 1)}{\underset{⏟}{ξ_{t}}} & \equiv [\begin{array}{c} y_{t} - μ \\ y_{t - 1} - μ \\ ⋮ \\ y_{t - p + 1} - μ \end{array}] \\ (10.1.10) & F_{(n p \times n p)} & \equiv [\begin{array}{c} Φ_{1} & Φ_{2} & Φ_{3} & \dots & Φ_{p - 1} & Φ_{p} \\ I_{n} & 0 & 0 & \dots & 0 & 0 \\ 0 & I_{n} & 0 & \dots & 0 & 0 \\ ⋮ & ⋮ & ⋮ & \dots & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & I_{n} & 0 \end{array}] \\ \underset{(n p \times 1)}{\underset{⏟}{v_{t}}} & \equiv [\begin{array}{c} ε_{t} \\ 0 \\ ⋮ \\ 0 \end{array}] . \end{aligned}

$VAR(p)$ $VAR(1)$ 形式：

\begin{matrix} (10.1.11) & ξ_{t} = F ξ_{t - 1} + v_{t}, \end{matrix}

其中

\begin{aligned} E (v_{t} v_{τ}^{'}) & = {\begin{cases} Q & for t = τ \\ 0 & otherwise \end{cases} \\ \underset{(n p \times n p)}{\underset{⏟}{Q}} & \equiv [\begin{array}{c} Ω & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 \\ ⋮ & ⋮ & \dots & ⋮ \\ 0 & 0 & \dots & 0 \end{array}] . \end{aligned}

Conditions for Stationarity

由 [10.1.11] 可得

\begin{matrix} (10.1.12) & ξ_{t + s} = v_{t + s} + F v_{t + s - 1} + F^{2} v_{t + s - 2} + \dots + F^{s - 1} v_{t + 1} + F^{s} ξ_{t} . \end{matrix}

$\boldsymbol{\varepsilon}_{t}$ $\mathbf{F}$ $VAR$ 是协方差平稳的。

下面的结果推广了 Chapter 1 中的命题 1.1。

Important

命题 10.1 ： $\mathbf{F}$ 的特征值满足

\begin{matrix} (10.1.13) & | I_{n} λ^{p} - Φ_{1} λ^{p - 1} - Φ_{2} λ^{p - 2} - \dots - Φ_{p} | = 0. \end{matrix}

$\lambda$ $\left| \lambda \right| < 1$ $VAR(p)$ 是协方差平稳的。等价地，若满足

| I_{n} - Φ_{1} z - Φ_{2} z^{2} - \dots - Φ_{p} z^{p} | = 0

$z$ $VAR$ 是协方差平稳的。

证明： $\mathbf{F} - \lambda \mathbf{I}_{np}$ 即可，不再赘述。

$\mathrm{MA}\left( \infty \right)$ Representation

$n$ 行构成了 [4.2.20] 的向量推广：

\begin{aligned} y_{t + s} & = μ + ε_{t + s} + Ψ_{1} ε_{t + s - 1} + Ψ_{2} ε_{t + s - 2} + \dots + Ψ_{s - 1} ε_{t + 1} \\ (10.1.14) & + F_{11}^{(s)} (y_{t} - μ) + F_{12}^{(s)} (y_{t - 1} - μ) + \dots + F_{1 p}^{(s)} (y_{t - p + 1} - μ) . \end{aligned}

${\boldsymbol{\Psi }}_{j} = {\mathbf{F}}_{11}^{\left( j\right) }$ ${\mathbf{F}}_{11}^{\left( j\right) }$ ${\mathbf{F}}^{j}$ ${\mathbf{F}}^{j}$ $\mathbf{F}$ $j$ $\left( {n \times n}\right)$ ${\mathbf{F}}_{11}^{\left( j\right) }$ $\left( {{np} \times {np}}\right)$ ${\mathbf{F}}^{j}$ $n$ $n$ ${\mathbf{F}}_{12}^{\left( j\right) }$ ${\mathbf{F}}^{j}$ $n$ $\left( {n + 1}\right)$ ${2n}$ ${\mathbf{F}}_{1p}^{\left( j\right) }$ ${\mathbf{F}}^{j}$ $n$ $\left\lbrack {n\left( {p - 1}\right) + 1}\right\rbrack$ ${np}$ 列。

$\mathbf{F}$ $s \rightarrow \infty$ ${\mathbf{F}}^{s} \rightarrow \mathbf{0}$ ${\mathbf{y}}_{t}$ $\boldsymbol{\varepsilon}$ 历史的收敛和：

\begin{matrix} (10.1.15) & y_{t} = μ + ε_{t} + Ψ_{1} ε_{t - 1} + Ψ_{2} ε_{t - 2} + Ψ_{3} ε_{t - 3} + \dots \equiv μ + Ψ (L) ε_{t}, \end{matrix}

$MA\left( \infty \right)$ 表示。

${\mathbf{y}}_{t - j}$ ${\boldsymbol{\varepsilon }}_{t - j},{\boldsymbol{\varepsilon }}_{t - j - 1},\ldots$ ${\boldsymbol{\varepsilon }}_{t + 1}$ $j = 0,1,\ldots$ ${\boldsymbol{\varepsilon }}_{t + 1}$ $j \geq 0$ ${\mathbf{y}}_{t - j}$ ${\mathbf{y}}_{t},{\mathbf{y}}_{t - 1},\ldots$ ${\mathbf{y}}_{t + 1}$ 的线性预测为

{\hat{y}}_{t + 1 ∣ t} = μ + Φ_{1} (y_{t} - μ) + Φ_{2} (y_{t - 1} - μ) + \dots + Φ_{p} (y_{t - p + 1} - μ),

${\boldsymbol{\varepsilon }}_{t + 1}$ ${y}_{t + 1}$ ${\mathbf{y}}_{t},{\mathbf{y}}_{t - 1}$ ${\mathbf{y}}_{t + 1}$ ${\mathbf{y}}_{t},{\mathbf{y}}_{t - 1}$ ${\mathbf{y}}_{t + s}$ 的预测形式为

\begin{aligned} (10.1.16) & {\hat{y}}_{t + s ∣ t} & = μ + F_{11}^{(s)} (y_{t} - μ) + F_{12}^{(s)} (y_{t - 1} - μ) + \dots + F_{1 p}^{(s)} (y_{t - p + 1} - μ) . \end{aligned}

${\boldsymbol{\Psi }}_{i}$ $\boldsymbol{\Phi }\left( L\right)$ $\boldsymbol{\Psi }\left( L\right)$ 的关系为

Ψ (L) = {[Φ (L)]}^{- 1},

这要求

[I_{n} - Φ_{1} L - Φ_{2} L^{2} - \dots - Φ_{p} L^{p}] [I_{n} + Ψ_{1} L + Ψ_{2} L^{2} + \dots] = I_{n} .

${L}^{1}$ 的系数设为零矩阵，得到

\begin{matrix} (10.1.17) & Ψ_{1} - Φ_{1} = 0. \end{matrix}

${L}^{2}$ 的系数设为零得到

\begin{matrix} (10.1.18) & Ψ_{2} = Φ_{1} Ψ_{1} + Φ_{2}, \end{matrix}

${L}^{s}$ 有

\begin{matrix} (10.1.19) & Ψ_{s} = Φ_{1} Ψ_{s - 1} + Φ_{2} Ψ_{s - 2} + \dots + Φ_{p} Ψ_{s - p} for s = 1, 2, \dots, \end{matrix}

${\boldsymbol{\Psi }}_{0} = {\mathbf{I}}_{n}$ $s < 0$ ${\boldsymbol{\Psi }}_{s} = \mathbf{0}$ 。

$MA\left( \infty \right)$ ${\boldsymbol{\varepsilon }}_{t}$ $\mathbf{y}$ ${\boldsymbol{\varepsilon }}_{t}$ $\mathbf{H}$ $\left( {n \times n}\right)$ 矩阵，定义

\begin{matrix} (10.1.20) & u_{t} \equiv H ε_{t} . \end{matrix}

$\mathbf{u}_{t}$ 当然是白噪声。此外，由 [10.1.15] 可写

\begin{aligned} (10.1.21) & y_{t} & = μ + H^{- 1} H ε_{t} + Ψ_{1} H^{- 1} H ε_{t - 1} + Ψ_{2} H^{- 1} H ε_{t - 2} + Ψ_{3} H^{- 1} H ε_{t - 3} + \dots \\ = μ + J_{0} u_{t} + J_{1} u_{t - 1} + J_{2} u_{t - 2} + J_{3} u_{t - 3} + \dots, \end{aligned}

其中

J_{s} \equiv Ψ_{s} H^{- 1} .

$\mathbf{H}$ ${\boldsymbol{\varepsilon }}_{t}$ $\boldsymbol{\Omega }$ 的矩阵：

H Ω H^{'} = D,

$\mathbf{D}$ $\mathbf{H}$ $\mathbf{u}_{t}$ 的元素彼此不相关：

E (u_{t} u_{t}^{'}) = E (H ε_{t} ε_{t}^{'} H^{'}) = D .

$VAR\left( p\right)$ $\mathbf{u}_{t}$ 的收敛无穷移动平均。

$MA\left( \infty \right)$ $MA$ $\left( {\boldsymbol{\Psi }}_{0}\right)$ $MA$ $\left( {\mathbf{J}}_{0}\right)$ $MA$ ${\boldsymbol{\Psi }}_{0} = {\mathbf{I}}_{n}$ 。

Assumptions Implicit in a VAR

$c$ ${\boldsymbol{\Phi }}_{1},\ldots ,{\boldsymbol{\Phi }}_{p}$ ${y}_{t}$ ${\mathbf{y}}_{t - 1},\ldots ,{\mathbf{y}}_{t - p}$ ${\boldsymbol{\Phi }}_{1},\ldots ,{\boldsymbol{\Phi }}_{p}$ ${\boldsymbol{\varepsilon }}_{t}$ ${\mathbf{y}}_{t - 1},\ldots ,{\mathbf{y}}_{t - p}$ $n$ $VAR$ ${\boldsymbol{\varepsilon }}_{t}$ ${\mathbf{y}}_{t - p - 1},{\mathbf{y}}_{t - p - 2},\ldots$ $\mathbf{y}$ $p$ $\mathbf{y}$ 各元素之间的所有动态相关性。

10.2 Autocovariances and Convergence Results for Vector Processes

$j$ th Autocovariance Matrix

$n$ $j$ $\left( {n \times n}\right)$ 矩阵：

\begin{matrix} (10.2.1) & Γ_{j} = E [(y_{t} - μ) {(y_{t - j} - μ)}^{'}] . \end{matrix}

${\gamma }_{j} = {\gamma }_{-j}$ ，但对于向量过程这并不成立：

Γ_{j} \neq Γ_{- j} .

${\boldsymbol{\Gamma }}_{j}$ $\left( {1,2}\right)$ ${y}_{1t}$ ${y}_{2, t - j}$ ${\boldsymbol{\Gamma }}_{-j}$ $\left( {1,2}\right)$ ${y}_{1t}$ ${y}_{2, t + j}$ ${y}_{1}$ ${y}_{2}$ ${y}_{2}$ ${y}_{1}$ 先前变动的响应完全不同。

正确的关系是

\begin{matrix} (10.2.2) & Γ_{j}^{'} = Γ_{- j} . \end{matrix}

$t$ $t + j$ ：

\begin{aligned} Γ_{j} & = E [(y_{t + j} - μ) {(y_{(t + j) - j} - μ)}^{'}] = E [(y_{t + j} - μ) {(y_{t} - μ)}^{'}] . \end{aligned}

取转置得到

Γ_{j}^{'} = E [(y_{t} - μ) {(y_{t + j} - μ)}^{'}] = Γ_{- j},

如所声称的。

$MA(q)$ Process

$q$ 阶向量移动平均过程的形式为

\begin{matrix} (10.2.3) & y_{t} = μ + ε_{t} + Θ_{1} ε_{t - 1} + Θ_{2} ε_{t - 2} + \dots + Θ_{q} ε_{t - q}, \end{matrix}

${\boldsymbol{\varepsilon }}_{t}$ ${\boldsymbol{\Theta }}_{i}$ $j = 1,2,\ldots , q$ $\left( {n \times n}\right)$ $MA$ ${\mathbf{y}}_{t}$ $\boldsymbol{\mu }$ ，方差为

\begin{aligned} (10.2.4) & Γ_{0} & = E [(y_{t} - μ) {(y_{t} - μ)}^{'}] \\ = E [ε_{t} ε_{t}^{'}] + Θ_{1} E [ε_{t - 1} ε_{t - 1}^{'}] Θ_{1}^{'} + Θ_{2} E [ε_{t - 2} ε_{t - 2}^{'}] Θ_{2}^{'} + \dots + Θ_{q} E [ε_{t - q} ε_{t - q}^{'}] Θ_{q}^{'} \\ = Ω + Θ_{1} Ω Θ_{1}^{'} + Θ_{2} Ω Θ_{2}^{'} + \dots + Θ_{q} Ω Θ_{q}^{'}, \end{aligned}

自协方差为

\begin{matrix} (10.2.5) & Γ_{j} = \sum_{v = 0}^{q - j} Θ_{j + v} Ω Θ_{v}^{'} for j = 1, 2, \dots, q, \end{matrix}

${\boldsymbol{\Theta }}_{0} = {\mathbf{I}}_{n}$ $MA\left( q\right)$ 过程都是协方差平稳的。

$MA(\infty)$ Process

$MA\left( \infty \right)$ 过程写为

\begin{matrix} (10.2.6) & y_{t} = μ + ε_{t} + Ψ_{1} ε_{t - 1} + Ψ_{2} ε_{t - 2} + \dots \end{matrix}

$\boldsymbol{\varepsilon}_t$ 也满足 [10.1.5] 和 [10.1.6]。

${\left\{ {h}_{s}\right\} }_{s = - \infty }^{\infty }$ ${\sum }_{s = - \infty }^{\infty }\left| {h}_{s}\right| < \infty$ $\left( {n \times m}\right)$ ${\mathbf{H}}_{s}$ ${\left\{ {\mathbf{H}}_{s}\right\} }_{s = - \infty }^{\infty }$ ${\psi }_{ij}^{\left( s\right) }$ $s$ ${\boldsymbol{\Psi }}_{s}$ $i$ $j$ ${\left\{ {\boldsymbol{\Psi }}_{s}\right\} }_{s = 0}^{\infty }$ 是绝对可和的，如果

\begin{matrix} (10.2.7) & \sum_{s = 0}^{\infty} | ψ_{i j}^{(s)} | < \infty for i = 1, 2, \dots, n and j = 1, 2, \dots, n . \end{matrix}

$MA\left( \infty \right)$ 过程的结果对向量过程也成立，因此我们有下面的命题。

Important

命题 10.2 ： ${\mathbf{y}}_{t}$ 是满足

y_{t} = μ + \sum_{k = 0}^{\infty} Ψ_{k} ε_{t - k}

$\left( {n \times 1}\right)$ ${\boldsymbol{\varepsilon }}_{t}$ ${\left\{ {\boldsymbol{\Psi }}_{k}\right\} }_{k = 0}^{\infty }$ ${y}_{it}$ ${\mathbf{y}}_{t}$ $i$ ${\mu }_{i}$ $\boldsymbol{\mu }$ $i$ 个元素。则

$i$ $t$ $j$ $s$ $E\left( {{y}_{it} - {\mu }_{i}}\right) \left( {{y}_{j, t - s} - {\mu }_{j}}\right)$ 存在，并由

Γ_{s} = \sum_{v = 0}^{\infty} Ψ_{s + v} Ω Ψ_{v}^{'} for s = 0, 1, 2, \dots;

$i$ $j$ 列元素给出；

${\left\{ {\boldsymbol{\Gamma }}_{s}\right\} }_{s = 0}^{\infty }$ 是绝对可和的。

${\left\{ {\boldsymbol{\varepsilon }}_{t}\right\} }_{t = - \infty }^{\infty }$ ${i}_{1},{i}_{2}$ ${i}_{3},{i}_{4} = 1,2,\ldots , n$ $E\left| {{\varepsilon }_{{i}_{1}, t}{\varepsilon }_{{i}_{2}, t}{\varepsilon }_{{i}_{3}, t}{\varepsilon }_{{i}_{4}, t}}\right| < \infty$ ，则还有

${i}_{1},{i}_{2},{i}_{3},{i}_{4} = 1,2,\ldots , n$ ${t}_{1},{t}_{2}$ ${t}_{3},{t}_{4}$ $E\left| {{y}_{{i}_{1},{t}_{1}}{y}_{{i}_{2},{t}_{2}}{y}_{{i}_{3},{t}_{3}}{y}_{{i}_{4},{t}_{4}}}\right| < \infty$ ；

$i, j = 1,2,\ldots , n$ $s$ $\left( {1/T}\right) \sum_{t = 1}^{T}{y}_{it}{y}_{j, t - s}\overset{p}{ \rightarrow }E\left( {{y}_{it}{y}_{j, t - s}}\right)$ 。

证明：

${z}_{t}\left( {i, l}\right)$ ${y}_{it}$ $\boldsymbol{\varepsilon}$ $l$ 个元素累积效应的分量：

\begin{matrix} (10.A.4) & z_{t} (i, l) = ψ_{i l}^{(0)} ε_{l t} + ψ_{i l}^{(1)} ε_{l, t - 1} + ψ_{i l}^{(2)} ε_{l, t - 2} + \dots = \sum_{v = 0}^{\infty} ψ_{i l}^{(v)} ε_{l, t - v}, \end{matrix}

${\psi }_{ij}^{\left( v\right) }$ ${\boldsymbol{\Psi }}_{v}$ $i$ $j$ $i$ ${y}_{it}$ $\boldsymbol{\varepsilon}$ $l = 1,2,\ldots , n$ 个分量各自贡献的和：

\begin{matrix} (10.A.5) & y_{i t} = μ_{i} + \sum_{l = 1}^{n} z_{t} (i, l) . \end{matrix}

${z}_{t}\left( {i, l}\right)$ ${\mathbf{y}}_{t}$ ${z}_{t}\left( {i, l}\right)$ 的这些表达式的有限和得到的。

证明 (a)： ${z}_{t}\left( {i, l}\right) \cdot {z}_{t - s}\left( {j, m}\right)$ $i, l, j$ $m$ $n$ $s$ $y$ 的自协方差的阶。由 [10.A.4] 注意到

\begin{aligned} (10.A.6) & E {z_{t} (i, l) \cdot z_{t - s} (j, m)} & = E {[\sum_{r = 0}^{\infty} ψ_{i l}^{(r)} ε_{l, t - r}] \times [\sum_{v = 0}^{\infty} ψ_{j m}^{(v)} ε_{m, t - s - v}]} \\ = \sum_{r = 0}^{\infty} \sum_{v = 0}^{\infty} {ψ_{i l}^{(r)} ψ_{j m}^{(v)}} \cdot E {ε_{l, t - r} ε_{m, t - s - v}} . \end{aligned}

上式成立是因为（即期望与求和可交换）

\sum_{r = 0}^{\infty} \sum_{v = 0}^{\infty} | ψ_{i l}^{(r)} ψ_{j m}^{(v)} | = \sum_{r = 0}^{\infty} \sum_{v = 0}^{\infty} | ψ_{i l}^{(r)} | \cdot | ψ_{j m}^{(v)} | = {\sum_{r = 0}^{\infty} | ψ_{i l}^{(r)} |} \times {\sum_{v = 0}^{\infty} | ψ_{j m}^{(v)} |} < \infty .

$\boldsymbol{\varepsilon}$ $\boldsymbol{\varepsilon}$ $r = s + v$ $r$ $r = s + v$ 处的值对这个和有贡献：

\begin{matrix} (10.A.7) & E {z_{t} (i, l) \cdot z_{t - s} (j, m)} = \sum_{v = 0}^{\infty} {ψ_{i l}^{(s + v)} ψ_{j m}^{(v)}} \cdot E {ε_{l, t - s - v} ε_{m, t - s - v}} = \sum_{v = 0}^{\infty} ψ_{i l}^{(s + v)} ψ_{j m}^{(v)} σ_{l m}, \end{matrix}

${\sigma }_{lm}$ ${\varepsilon }_{lt}$ ${\varepsilon }_{mt}$ $\boldsymbol{\Omega }$ $l$ $m$ 列元素给出。

${\boldsymbol{\Gamma }}_{s}$ $i$ $j$ 列元素给出

γ_{i j}^{(s)} = E (y_{i t} - μ_{i}) (y_{j, t - s} - μ_{j}) .

使用 [10.A.5] 和 [10.A.7]，这可以表示为

\begin{aligned} (10.A.8) & E (y_{i t} - μ_{i}) (y_{j, t - s} - μ_{j}) & = E {[\sum_{l = 1}^{n} z_{t} (i, l)] [\sum_{m = 1}^{n} z_{t - s} (j, m)]} \\ = \sum_{l = 1}^{n} \sum_{m = 1}^{n} E {z_{t} (i, l) \cdot z_{t - s} (j, m)} \\ = \sum_{l = 1}^{n} \sum_{m = 1}^{n} \sum_{v = 0}^{\infty} ψ_{i l}^{(s + v)} ψ_{j m}^{(v)} σ_{l m} \\ = \sum_{v = 0}^{\infty} \sum_{l = 1}^{n} \sum_{m = 1}^{n} ψ_{i l}^{(s + v)} ψ_{j m}^{(v)} σ_{l m} . \end{aligned}

$\sum_{l = 1}^{n}\sum_{m = 1}^{n}{\psi }_{il}^{\left( s + v\right) }{\psi }_{jm}^{\left( v\right) }{\sigma }_{lm}$ ${\boldsymbol{\Psi }}_{s + v}\boldsymbol{\Omega }{\boldsymbol{\Psi }}_{v}^{\prime }$ $i$ $j$ ${\boldsymbol{\Gamma }}_{s}$ $i$ $j$ $\sum_{v = 0}^{\infty }{\boldsymbol{\Psi }}_{s + v}\boldsymbol{\Omega }{\boldsymbol{\Psi }}_{v}^{\prime }$ $i$ $j$ 列元素给出，如 (a) 部分所声称的。

证明 (b)： ${h}_{s}\left( {i, j, l, m}\right)$ 为 [10.A.7] 中的矩：

h_{s} (i, j, l, m) \equiv E {z_{t} (i, l) \cdot z_{t - s} (j, m)} = \sum_{v = 0}^{\infty} ψ_{i l}^{(s + v)} ψ_{j m}^{(v)} σ_{l m};

${\left\{ {h}_{s}\left( \cdot \right) \right\} }_{s = 0}^{\infty }$ 是绝对可和的：

\begin{aligned} (10.A.9) & \sum_{s = 0}^{\infty} | h_{s} (i, j, l, m) | & \leq \sum_{s = 0}^{\infty} \sum_{v = 0}^{\infty} | ψ_{i l}^{(s + v)} | \cdot | ψ_{j m}^{(v)} | \cdot | σ_{l m} | \\ = | σ_{l m} | \sum_{v = 0}^{\infty} | ψ_{j m}^{(v)} | \sum_{s = 0}^{\infty} | ψ_{i l}^{(s + v)} | \\ \leq | σ_{l m} | \sum_{v = 0}^{\infty} | ψ_{j m}^{(v)} | \sum_{v = 0}^{\infty} | ψ_{i l}^{(v)} | \\ < \infty . \end{aligned}

${\boldsymbol{\Gamma }}_{s}$ $i$ $j$ 列元素在 [10.A.8] 中被看到由

γ_{i j}^{(s)} = \sum_{l = 1}^{n} \sum_{m = 1}^{n} h_{s} (i, j, l, m) .

给出。因此，

\begin{matrix} (10.A.10) & \sum_{s = 0}^{\infty} | γ_{i j}^{(s)} | \leq \sum_{s = 0}^{\infty} \sum_{l = 1}^{n} \sum_{m = 1}^{n} | h_{s} (i, j, l, m) | = \sum_{l = 1}^{n} \sum_{m = 1}^{n} \sum_{s = 0}^{\infty} | h_{s} (i, j, l, m) | . \end{matrix}

$M < \infty$ 使得

\sum_{s = 0}^{\infty} | h_{s} (i, j, l, m) | < M

$i, j, l$ $m$ 的值。因此，[10.A.10] 意味着

\sum_{s = 0}^{\infty} | γ_{i j}^{(s)} | < \sum_{l = 1}^{n} \sum_{m = 1}^{n} M = n^{2} M < \infty,

${\left\{ {\boldsymbol{\Gamma }}_{s}\right\} }_{s = 0}^{\infty }$ $i$ $j$ 列元素是绝对可和的，如 (b) 部分所声称的。

证明 (c)： 本质上与命题 7.10 的证明相同的代数运算建立了

\begin{aligned} (10.A.11) & E | z_{t_{1}} (i_{1}, l_{1}) \cdot z_{t_{2}} (i_{2}, l_{2}) \cdot z_{t_{3}} (i_{3}, l_{3}) \cdot z_{t_{4}} (i_{4}, l_{4}) | & = E {{\sum_{v_{1} = 0}^{\infty} ψ_{i_{1}, l_{1}}^{(v_{1})} ε_{l_{1}, t_{1} - v_{1}}} \cdot {\sum_{v_{2} = 0}^{\infty} ψ_{i_{2}, l_{2}}^{(v_{2})} ε_{l_{2}, t_{2} - v_{2}}} \\ \cdot {\sum_{v_{3} = 0}^{\infty} ψ_{i_{3}, l_{3}}^{(v_{3})} ε_{l_{3}, t_{3} - v_{3}}} \cdot {\sum_{v_{4} = 0}^{\infty} ψ_{i_{4}, l_{4}}^{(v_{4})} ε_{l_{4}, t_{4} - v_{4}}} | \\ \leq \sum_{v_{1} = 0}^{\infty} \sum_{v_{2} = 0}^{\infty} \sum_{v_{3} = 0}^{\infty} \sum_{v_{4} = 0}^{\infty} | ψ_{i_{1}, l_{1}}^{(v_{1})} ψ_{i_{2}, l_{2}}^{(v_{2})} ψ_{i_{3}, l_{3}}^{(v_{3})} ψ_{i_{4}, l_{4}}^{(v_{4})} | \\ \times E | ε_{l_{1}, t_{1} - v_{1}} ε_{l_{2}, t_{2} - v_{2}} ε_{l_{3}, t_{3} - v_{3}} ε_{l_{4}, t_{4} - v_{4}} | \\ < \infty . \end{aligned}

现在，

\begin{aligned} E | y_{i_{1}, t_{1}} y_{i_{2}, t_{2}} y_{i_{3}, t_{3}} y_{i_{4}, t_{4}} | & = E | μ_{i_{1}} + \sum_{l_{1} = 1}^{n} z_{t_{1}} (i_{1}, l_{1}) | \cdot | μ_{i_{2}} + \sum_{l_{2} = 1}^{n} z_{t_{2}} (i_{2}, l_{2}) | \\ \cdot | μ_{i_{3}} + \sum_{l_{3} = 1}^{n} z_{t_{3}} (i_{3}, l_{3}) | \cdot | μ_{i_{4}} + \sum_{l_{4} = 1}^{n} z_{t_{4}} (i_{4}, l_{4}) | \\ \leq E {| μ_{i_{1}} | + \sum_{l_{1} = 1}^{n} | z_{t_{1}} (i_{1}, l_{1}) |} \cdot {| μ_{i_{2}} | + \sum_{l_{2} = 1}^{n} | z_{t_{2}} (i_{2}, l_{2}) |} \\ \cdot {| μ_{i_{3}} | + \sum_{l_{3} = 1}^{n} | z_{t_{3}} (i_{3}, l_{3}) |} \cdot {| μ_{i_{4}} | + \sum_{l_{4} = 1}^{n} | z_{t_{4}} (i_{4}, l_{4}) |} . \end{aligned}

$z$ 的一到三阶矩的项，这些也必须有限。

证明 (d)： 注意到

z_{t} (i, l) \cdot z_{t - s} (j, m) = \sum_{r = 0}^{\infty} \sum_{v = 0}^{\infty} ψ_{i l}^{(r)} ψ_{j m}^{(v)} ε_{l, t - r} ε_{m, t - s - v} .

导致 [7.2.14] 的相同论证可用于建立

\begin{matrix} (10.A.12) & (1 / T) \sum_{t = 1}^{T} z_{t} (i, l) \cdot z_{t - s} (j, m) \overset{p}{\to} E {z_{t} (i, l) \cdot z_{t - s} (j, m)} . \end{matrix}

$y$ 的二阶矩的遍历性，由 [10.A.5] 注意到

\begin{aligned} (1 / T) \sum_{t = 1}^{T} y_{i t} y_{j, t - s} & = (1 / T) \sum_{t = 1}^{T} [μ_{i} + \sum_{l = 1}^{n} z_{t} (i, l)] [μ_{j} + \sum_{m = 1}^{n} z_{t - s} (j, m)] \\ = μ_{i} μ_{j} + μ_{i} \sum_{m = 1}^{n} [(1 / T) \sum_{t = 1}^{T} z_{t - s} (j, m)] + μ_{j} \sum_{l = 1}^{n} [(1 / T) \sum_{t = 1}^{T} z_{t} (i, l)] \\ + \sum_{l = 1}^{n} \sum_{m = 1}^{n} [(1 / T) \sum_{t = 1}^{T} z_{t} (i, l) z_{t - s} (j, m)] \\ \overset{p}{\to} μ_{i} μ_{j} + μ_{i} \sum_{m = 1}^{n} E [z_{t - s} (j, m)] + μ_{j} \sum_{l = 1}^{n} E [z_{t} (i, l)] \\ + \sum_{l = 1}^{n} \sum_{m = 1}^{n} E [z_{t} (i, l) z_{t - s} (j, m)] \\ = E {[μ_{i} + \sum_{l = 1}^{n} z_{t} (i, l)] [μ_{j} + \sum_{m = 1}^{n} z_{t - s} (j, m)]} \\ = E [y_{i t} y_{j, t - s}], \end{aligned}

如所声称的。证毕。

$MA\left( \infty \right)$ $q \rightarrow \infty$ ${\mathbf{y}}_{t}$ ${\mathbf{y}}_{t}$ 对二阶矩是遍历的。

$MA\left( \infty \right)$ ${\boldsymbol{\Psi }}_{s}$ ${\mathbf{F}}^{s}$ $\mathbf{F}$ ${np}$ $\left( {{\lambda }_{1},{\lambda }_{2},\ldots ,{\lambda }_{np}}\right)$ ${\boldsymbol{\Psi }}_{s}$ 的任何元素都可以写为这些特征值的加权平均，如 [1.2.20]：

ψ_{i j}^{(s)} = c_{1} (i, j) \cdot λ_{1}^{s} + c_{2} (i, j) \cdot λ_{2}^{s} + \dots + c_{n p} (i, j) \cdot λ_{n p}^{s},

${c}_{v}\left( {i, j}\right)$ $v, i$ $j$ $s$ 的常数。

Multivariate Filters

$\left( {n \times 1}\right)$ ${\mathbf{y}}_{t}$ $MA\left( \infty \right)$ 过程：

\begin{matrix} (10.2.8) & y_{t} = μ_{Y} + Ψ (L) ε_{t}, \end{matrix}

${\left\{ {\boldsymbol{\Psi }}_{k}\right\} }_{k = 0}^{\infty }$ ${\left\{ {\mathbf{H}}_{k}\right\} }_{k = - \infty }^{\infty }$ $\left( {r \times n}\right)$ $\left( {r \times 1}\right)$ ${\mathbf{x}}_{t}$ ${\mathbf{y}}_{t}$ 的关系为

\begin{matrix} (10.2.9) & x_{t} = H (L) y_{t} = \sum_{k = - \infty}^{\infty} H_{k} y_{t - k} . \end{matrix}

即

\begin{aligned} x_{t} & = H (L) [μ_{Y} + Ψ (L) ε_{t}] \\ (10.2.10) & = H (1) μ_{Y} + H (L) Ψ (L) ε_{t} \\ = μ_{x} + B (L) ε_{t}, \end{aligned}

${\boldsymbol{\mu }}_{\mathbf{x}} \equiv \mathbf{H}\left( 1\right) {\boldsymbol{\mu }}_{\mathbf{Y}}$ $\mathbf{B}\left( L\right)$ 是由

\begin{matrix} (10.2.11) & B (L) = \sum_{k = - \infty}^{\infty} B_{k} L^{k} = H (L) Ψ (L) \end{matrix}

给出的复合算子。

${\mathbf{x}}_{t}$ $MA\left( \infty \right)$ 过程。

Important

命题 10.3 ： ${\left\{ {\boldsymbol{\Psi }}_{k}\right\} }_{k = 0}^{\infty }$ $\left( {n \times n}\right)$ ${\left\{ {\mathbf{H}}_{k}\right\} }_{k = - \infty }^{\infty }$ $\left( {r \times n}\right)$ $\mathbf{B}\left( L\right) = \mathbf{H}\left( L\right) \boldsymbol{\Psi }\left( L\right)$ ${\left\{ {\mathbf{B}}_{k}\right\} }_{k = - \infty }^{\infty }$ 是绝对可和的。

$\left\{ {\boldsymbol{\varepsilon }}_{t}\right\}$ $\left\{ {\mathbf{x}}_{t}\right\}$ 具有有限四阶矩且对二阶矩是遍历的。

证明：

将 [10.2.11] 明确写出：

\begin{aligned} H (L) Ψ (L) & = (\dots + H_{- 1} L^{- 1} + H_{0} L^{0} + H_{1} L^{1} + \dots) \times (Ψ_{0} L^{0} + Ψ_{1} L^{1} + Ψ_{2} L^{2} + \dots), \end{aligned}

${L}^{k}$ 的系数为

\begin{matrix} (10.A.13) & B_{k} = H_{k} Ψ_{0} + H_{k - 1} Ψ_{1} + H_{k - 2} Ψ_{2} + \dots . \end{matrix}

${b}_{ij}^{\left( k\right) }$ ${\mathbf{B}}_{k}$ $i$ $j$ ${h}_{ij}^{\left( k\right) }$ ${\psi }_{ij}^{\left( k\right) }$ ${\mathbf{H}}_{k}$ ${\boldsymbol{\Psi }}_{k}$ $i$ $j$ $i$ $j$ 列元素表明

b_{i j}^{(k)} = \sum_{m = 1}^{n} h_{i m}^{(k)} ψ_{m j}^{(0)} + \sum_{m = 1}^{n} h_{i m}^{(k - 1)} ψ_{m j}^{(1)} + \sum_{m = 1}^{n} h_{i m}^{(k - 2)} ψ_{m j}^{(2)} + \dots = \sum_{r = 0}^{\infty} \sum_{m = 1}^{n} h_{i m}^{(k - r)} ψ_{m j}^{(r)} .

因此，

\begin{aligned} (10.A.14) & \sum_{k = - \infty}^{\infty} | b_{i j}^{(k)} | & = \sum_{k = - \infty}^{\infty} | \sum_{r = 0}^{\infty} \sum_{m = 1}^{n} h_{i m}^{(k - r)} ψ_{m j}^{(r)} | \\ \leq \sum_{k = - \infty}^{\infty} \sum_{r = 0}^{\infty} \sum_{m = 1}^{n} | h_{i m}^{(k - r)} ψ_{m j}^{(r)} | \\ = \sum_{m = 1}^{n} \sum_{r = 0}^{\infty} | ψ_{m j}^{(r)} | \sum_{k = - \infty}^{\infty} | h_{i m}^{(k - r)} | \\ = \sum_{m = 1}^{n} \sum_{r = 0}^{\infty} | ψ_{m j}^{(r)} | \sum_{k = - \infty}^{\infty} | h_{i m}^{(k)} | . \end{aligned}

${\left\{ {\mathbf{H}}_{k}\right\} }_{k = - \infty }^{\infty }$ ${\left\{ {\boldsymbol{\Psi }}_{k}\right\} }_{k = 0}^{\infty }$ ${M}_{1}$ ${M}_{2}$ 使得

\begin{aligned} \sum_{k = - \infty}^{\infty} | h_{i m}^{(k)} | & < M_{1} < \infty \\ \sum_{r = 0}^{\infty} | ψ_{m j}^{(r)} | & < M_{2} < \infty . \end{aligned}

因此，[10.A.14] 变为

\sum_{k = - \infty}^{\infty} | b_{i j}^{(k)} | < \sum_{m = 1}^{n} M_{1} M_{2} = n M_{1} M_{2} < \infty,

${\left\{ {\mathbf{B}}_{k}\right\} }_{k = - \infty }^{\infty }$ 是绝对可和的。

$\left\{ {\boldsymbol{\varepsilon }}_{t}\right\}$ $\left\{ {\mathbf{x}}_{t}\right\}$ 具有有限四阶矩且对二阶矩是遍历的，这由命题 10.2 的结果 (c) 和 (d) 直接得出。证毕。

Vector Autoregression

$VAR\left( p\right)$ $\mathbf{y}$ ${\boldsymbol{\xi }}_{t}$ ${\boldsymbol{\xi }}_{t}$ $\mathbf{y}$ $\boldsymbol{\Sigma }$ ${\boldsymbol{\xi }}_{t}$ 的方差：

\begin{aligned} (10.2.12) & Σ & = E (ξ_{t} ξ_{t}^{'}) \\ = E {[\begin{array}{c} y_{t} - μ \\ y_{t - 1} - μ \\ ⋮ \\ y_{t - p + 1} - μ \end{array}] \times [{(y_{t} - μ)}^{'} {(y_{t - 1} - μ)}^{'} \dots {(y_{t - p + 1} - μ)}^{'}]} \\ = [\begin{array}{c} Γ_{0} & Γ_{1} & \dots & Γ_{p - 1} \\ Γ_{1}^{'} & Γ_{0} & \dots & Γ_{p - 2} \\ ⋮ & ⋮ & \dots & ⋮ \\ Γ_{p - 1}^{'} & Γ_{p - 2}^{'} & \dots & Γ_{0} \end{array}], \end{aligned}

${\boldsymbol{\Gamma }}_{j}$ $\mathbf{y}$ $j$ 个自协方差。将 [10.1.11] 右乘其转置并取期望得到

E [ξ_{t} ξ_{t}^{'}] = E [(F ξ_{t - 1} + v_{t}) {(F ξ_{t - 1} + v_{t})}^{'}] = F E (ξ_{t - 1} ξ_{t - 1}^{'}) F^{'} + E (v_{t} v_{t}^{'}) .

或

\begin{matrix} (10.2.13) & Σ = F Σ F^{'} + Q . \end{matrix}

$\mathbf{A}$ $\left( {m \times n}\right)$ $\operatorname{vec}\left( \mathbf{A}\right)$ $\left( {{mn} \times 1}\right)$ $\mathbf{A}$ 的列从左到右依次堆叠得到。例如，如果

\begin{matrix} A = [\begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{array}] \end{matrix}

则

\begin{matrix} (10.2.14) & \begin{matrix} vec (A) = [\begin{matrix} a_{11} \\ a_{21} \\ a_{31} \\ a_{12} \\ a_{22} \\ a_{32} \end{matrix}] . \end{matrix} \end{matrix}

Important

命题 10.4 ： $\mathbf{A},\mathbf{B}$ $\mathbf{C}$ 是维度使得乘积 ABC 存在的矩阵，则

\begin{matrix} (10.2.15) & vec (ABC) = (C^{'} \otimes A) \cdot vec (B), \end{matrix}

$\otimes$ 表示 Kronecker 积，也称张量积，并且有

\begin{matrix} A \otimes B = [\begin{matrix} a_{11} B & a_{12} B & \dots & a_{1 n} B \\ a_{21} B & a_{22} B & \dots & a_{2 n} B \\ ⋮ & ⋮ & ⋱ & ⋮ \\ a_{m 1} B & a_{m 2} B & \dots & a_{m n} B \end{matrix}] . \end{matrix}

证明：

$\mathbf{A}$ $\left( {m \times n}\right)$ $\mathbf{B}$ $\left( {n \times r}\right)$ $\mathbf{C}$ $\left( {r \times q}\right)$ $\left( {n \times 1}\right)$ ${\mathbf{b}}_{i}$ $\mathbf{B}$ $i$ ${c}_{ij}$ $\mathbf{C}$ $i$ $j$ 列元素。则

\begin{aligned} ABC & = A [\begin{array}{llll} b_{1} & b_{2} & \dots & b_{r} \end{array}] [\begin{array}{c} c_{11} & c_{12} & \dots & c_{1 q} \\ c_{21} & c_{22} & \dots & c_{2 q} \\ ⋮ & ⋮ & \dots & ⋮ \\ c_{r 1} & c_{r 2} & \dots & c_{r q} \end{array}] \\ = [{A b_{1} c_{11} + A b_{2} c_{21} + \dots + A b_{r} c_{r 1}} \\ {A b_{1} c_{12} + A b_{2} c_{22} + \dots + A b_{r} c_{r 2}} \dots \\ {A b_{1} c_{1 q} + A b_{2} c_{2 q} + \dots + A b_{r} c_{r q}}] \\ = [{c_{11} A b_{1} + c_{21} A b_{2} + \dots + c_{r 1} A b_{r}} \\ {c_{12} A b_{1} + c_{22} A b_{2} + \dots + c_{r 2} A b_{r}} \dots \\ {c_{1 q} A b_{1} + c_{2 q} A b_{2} + \dots + c_{r q} A b_{r}}] . \end{aligned}

应用 vec 算子得到

\begin{aligned} vec (ABC) & = [\begin{array}{c} c_{11} A b_{1} + c_{21} A b_{2} + \dots + c_{r 1} A b_{r} \\ c_{12} A b_{1} + c_{22} A b_{2} + \dots + c_{r 2} A b_{r} \\ ⋮ \\ c_{1 q} A b_{1} + c_{2 q} A b_{2} + \dots + c_{r q} A b_{r} \end{array}] \\ = [\begin{array}{c} c_{11} A & c_{21} A & \dots & c_{r 1} A \\ c_{12} A & c_{22} A & \dots & c_{r 2} A \\ ⋮ & ⋮ & \dots & ⋮ \\ c_{1 q} A & c_{2 q} A & \dots & c_{r q} A \end{array}] [\begin{array}{c} b_{1} \\ b_{2} \\ ⋮ \\ b_{r} \end{array}] \\ = (C^{'} \otimes A) \cdot vec (B) . \end{aligned}

证毕。

$AXB = C$ 可以转化为
$(B^{'} \otimes A) vec (X) = vec (C) .$

因此，如果将 vec 算子应用于 [10.2.13] 的两边，结果是

\begin{aligned} vec (Σ) & = (F \otimes F) \cdot vec (Σ) + vec (Q) \\ (10.2.16) & = A vec (Σ) + vec (Q), \end{aligned}

其中

\begin{matrix} (10.2.17) & A \equiv (F \otimes F) . \end{matrix}

$r = {np}$ $\mathbf{F}$ $\left( {r \times r}\right)$ $\mathcal{A}$ $\left( {{r}^{2} \times {r}^{2}}\right)$ 矩阵。[10.2.16] 的解为

\begin{matrix} (10.2.18) & vec (Σ) = {[I_{r^{2}} - A]}^{- 1} vec (Q), \end{matrix}

$\left\lbrack {{\mathbf{I}}_{{r}^{2}} - \mathcal{A}}\right\rbrack$ $\mathcal{A}$ $\mathbf{F} \otimes \mathbf{F}$ ${\lambda }_{i}{\lambda }_{j}$ ${\lambda }_{i}$ ${\lambda }_{j}$ $\mathbf{F}$ $i$ $\left| {\lambda }_{i}\right| < 1$ $\mathcal{A}$ $\left\lbrack {{\mathbf{I}}_{{r}^{2}} - \mathcal{A}}\right\rbrack$ 确实是非奇异的。

$VAR\left( p\right)$ $p$ 个自协方差矩阵可以通过将 [10.2.12] 代入 [10.2.18] 计算：

\begin{matrix} (10.2.19) & \begin{matrix} vec [\begin{matrix} Γ_{0} & Γ_{1} & \dots & Γ_{p - 1} \\ Γ_{1}^{'} & Γ_{0} & \dots & Γ_{p - 2} \\ ⋮ & ⋮ & \dots & ⋮ \\ Γ_{p - 1}^{'} & Γ_{p - 2}^{'} & \dots & Γ_{0} \end{matrix}] = {[I_{r^{2}} - A]}^{- 1} vec (Q) . \end{matrix} \end{matrix}

${\boldsymbol{\xi }}_{t}$ $j$ ${\boldsymbol{\Sigma }}_{j}$ ${\boldsymbol{\xi }}_{t - j}^{\prime }$ 并取期望得到：

E (ξ_{t} ξ_{t - j}^{'}) = F \cdot E (ξ_{t - 1} ξ_{t - j}^{'}) + E (v_{t} ξ_{t - j}^{'}) .

因此，

\begin{aligned} (10.2.20) & Σ_{j} & = F Σ_{j - 1} for j = 1, 2, \dots, \\ (10.2.21) & Σ_{j} & = F^{j} Σ for j = 1, 2, \dots . \end{aligned}

${\mathbf{y}}_{t}$ $j$ ${\boldsymbol{\Gamma }}_{j}$ $n$ $n$ ${\boldsymbol{\Sigma }}_{j} = \mathbf{F}{\boldsymbol{\Sigma }}_{j - 1}$ ${\boldsymbol{\Sigma }}_{j}$ 的结构为

\begin{matrix} Σ_{j} = [\begin{matrix} Γ_{j} & Γ_{j + 1} & \dots & Γ_{j + p - 1} \\ Γ_{j - 1}^{'} & Γ_{j} & \dots & Γ_{j + p - 2} \\ ⋮ & ⋮ & \dots & ⋮ \\ Γ_{j - p + 1}^{'} & Γ_{j - p + 2}^{'} & \dots & Γ_{j} \end{matrix}], \end{matrix}

$\mathbf{F}$ $\mathbf{F}{\boldsymbol{\Sigma }}_{j - 1}$ $n \times n$ ${\boldsymbol{\Gamma }}_{j}$ ）为

\begin{aligned} Γ_{j} & = [Φ_{1} Φ_{2} \dots Φ_{p}] [\begin{array}{c} Γ_{j - 1} \\ Γ_{j - 2} \\ ⋮ \\ Γ_{j - p} \end{array}] \\ = Φ_{1} Γ_{j - 1} + Φ_{2} Γ_{j - 2} + \dots + Φ_{p} Γ_{j - p} . \end{aligned}

因此得到递推关系

\begin{matrix} (10.2.22) & Γ_{j} = Φ_{1} Γ_{j - 1} + Φ_{2} Γ_{j - 2} + \dots + Φ_{p} Γ_{j - p} for j = p, p + 1, p + 2, \dots . \end{matrix}

$\left( {{\mathbf{y}}_{t - j} - \boldsymbol{\mu }}\right)^{\prime }$ ${\boldsymbol{\varepsilon }}_{t}$ ${\mathbf{y}}_{t - j}$ $j \geq 0$ ）不相关，即可得到相同的结果。

10.3 The Autocovariance-Generating Function for Vector Processes

Definition of Autocovariance-Generating Function for Vector Processes

${y}_{t}$ ${g}_{Y}\left( z\right)$ 定义为

g_{Y} (z) \equiv \sum_{j = - \infty}^{\infty} γ_{j} z^{j}

其中

γ_{j} \equiv E [(y_{t} - μ) (y_{t - j} - μ)]

$z$ ${\mathbf{y}}_{t}$ ${\mathbf{G}}_{\mathbf{Y}}\left( z\right)$ 定义为

\begin{matrix} (10.3.1) & G_{Y} (z) \equiv \sum_{j = - \infty}^{\infty} Γ_{j} z^{j} \end{matrix}

其中

Γ_{j} \equiv E [(y_{t} - μ) {(y_{t - j} - μ)}^{'}]

$z$ 同样是复标量。

Autocovariance-Generating Function for a Vector Moving Average Process

${\boldsymbol{\varepsilon }}_{t}$ ，其自协方差生成函数为

\begin{matrix} (10.3.2) & G_{ε} (z) = Ω . \end{matrix}

${MA}\left( q\right)$ 过程，单变量情形下式 [3.6.3] 的自协方差生成函数推广为

\begin{aligned} (10.3.3) & G_{Y} (z) & = (I_{n} + Θ_{1} z + Θ_{2} z^{2} + \dots + Θ_{q} z^{q}) Ω \\ \times (I_{n} + Θ_{1}^{'} z^{- 1} + Θ_{2}^{'} z^{- 2} + \dots + Θ_{q}^{'} z^{- q}) . \end{aligned}

${z}^{j}$ ${\boldsymbol{\Gamma }}_{j}$ 来验证。

对于形如

y_{t} = μ + Ψ_{0} ε_{t} + Ψ_{1} ε_{t - 1} + Ψ_{2} ε_{t - 2} + \dots = μ + Ψ (L) ε_{t}

${MA}\left( \infty \right)$ ${\left\{ {\boldsymbol{\Psi }}_{k}\right\} }_{k = 0}^{\infty }$ 绝对可和，式 [10.3.3] 推广为

\begin{matrix} (10.3.4) & G_{Y} (z) = [Ψ (z)] Ω {[Ψ (z^{- 1})]}^{'} . \end{matrix}

Autocovariance-Generating Function for a Vector Autoregression

${VAR}\left( 1\right)$ ${\boldsymbol{\xi }}_{t} = \mathbf{F}{\boldsymbol{\xi }}_{t - 1} + {\mathbf{v}}_{t}$ $\mathbf{F}$ $\boldsymbol{\xi }_{t}$ $\left( {r \times 1}\right)$ $E\left( {{\mathbf{v}}_{t}{\mathbf{v}}_{t}^{\prime }}\right) = \mathbf{Q}$ 。式 [10.3.4] 意味着自协方差生成函数可以表示为

\begin{aligned} (10.3.5) & G_{ξ} (z) & = {[I_{r} - F z]}^{- 1} Q {[I_{r} - F^{'} z^{- 1}]}^{- 1} \\ = [I_{r} + F z + F^{2} z^{2} + F^{3} z^{3} + \dots] Q \\ \times [I_{r} + (F^{'}) z^{- 1} + {(F^{'})}^{2} z^{- 2} + {(F^{'})}^{3} z^{- 3} + \dots] . \end{aligned}

Transformations of Vector Processes

两个互不相关的单变量过程之和的自协方差生成函数等于它们各自的自协方差生成函数之和（式 [4.7.19]）。这一结果可以很容易地推广到向量情形：

\begin{aligned} G_{X + W} (z) & = \sum_{j = - \infty}^{\infty} E [(x_{t} + w_{t} - μ_{X} - μ_{W}) \\ \times (x_{t - j} + w_{t - j} - μ_{X} - μ_{W})^{'}] z^{j} \\ = \sum_{j = - \infty}^{\infty} E [(x_{t} - μ_{X}) (x_{t - j} - μ_{X})^{'} z^{j}] \\ + \sum_{j = - \infty}^{\infty} E [(w_{t} - μ_{W}) (w_{t - j} - μ_{W})^{'} z^{j}] \\ = G_{X} (z) + G_{W} (z) . \end{aligned}

$\left( {r \times 1}\right)$ ${\boldsymbol{\xi }}_{t}$ $\left( {n \times r}\right)$ ${\mathbf{H}}^{\prime }$ ${\mathbf{H}}^{\prime }$ $\mathbf{H}$ ：

E [(H^{'} ξ_{t} - H^{'} μ_{ξ}) {(H^{'} ξ_{t - j} - H^{'} μ_{ξ})}^{'}] = H^{'} E [(ξ_{t} - μ_{ξ}) {(ξ_{t - j} - μ_{ξ})}^{'}] H,

这意味着

G_{H^{'} ξ} (z) = H^{'} G_{ξ} (z) H .

$r$ ${VAR}\left( 1\right)$ ${\boldsymbol{\xi }}_{t} = \mathbf{F}{\boldsymbol{\xi }}_{t - 1} + {\mathbf{v}}_{t}$ ${\mathbf{u}}_{t} = {\mathbf{H}}^{\prime }{\boldsymbol{\xi }}_{t} + {\mathbf{w}}_{t}$ ${\mathbf{u}}_{t}$ ${\mathbf{w}}_{t}$ $j$ ${\boldsymbol{\xi }}_{t - j}$ 都不相关的白噪声过程。则

G_{U} (z) = H^{'} G_{ξ} (z) H + G_{W} (z),

$\mathbf{R}$ ${\mathbf{w}}_{t}$ 的方差，则

\begin{matrix} (10.3.6) & G_{U} (z) = H^{'} {[I_{r} - F z]}^{- 1} Q {[I_{r} - F^{'} z^{- 1}]}^{- 1} H + R . \end{matrix}

更一般地，考虑由

y_{t} = μ_{Y} + Ψ (L) ε_{t}

$\left( {n \times 1}\right)$ ${\mathbf{y}}_{t}$ ${\boldsymbol{\varepsilon }}_{t}$ $\boldsymbol{\Omega }$ $\boldsymbol{\Psi }\left( L\right) = \sum_{k = 0}^{\infty }{\boldsymbol{\Psi }}_{k}{L}^{k}$ ${\left\{ {\boldsymbol{\Psi }}_{k}\right\} }_{k = 0}^{\infty }$ $\mathbf{y}$ 的自协方差生成函数为

\begin{matrix} (10.3.7) & G_{Y} (z) = Ψ (z) Ω {[Ψ (z^{- 1})]}^{'} . \end{matrix}

${\left\{ {\mathbf{H}}_{k}\right\} }_{k = - \infty }^{\infty }$ $\left( {r \times n}\right)$ $\left( {r \times 1}\right)$ ${\mathbf{x}}_{t}$ ${\mathbf{y}}_{t}$ 按照

x_{t} = H (L) y_{t} = \sum_{k = - \infty}^{\infty} H_{k} y_{t - k} = μ_{x} + B (L) ε_{t}

${\boldsymbol{\mu }}_{\mathbf{X}} = \mathbf{H}\left( 1\right) {\boldsymbol{\mu }}_{\mathbf{Y}}$ $\mathbf{B}\left( L\right) = \mathbf{H}\left( L\right) \boldsymbol{\Psi }\left( L\right)$ $\mathbf{x}$ 的自协方差生成函数可以从下式得到：

\begin{aligned} (10.3.8) & G_{X} (z) & = B (z) Ω {[B (z^{- 1})]}^{'} \\ = [H (z) Ψ (z)] Ω {[Ψ (z^{- 1})]}^{'} {[H (z^{- 1})]}^{'} . \end{aligned}

${\mathbf{y}}_{t}$ $\mathbf{H}\left( L\right)$ $\mathbf{H}\left( z\right)$ ${\mathbf{H}\left( {z}^{-1}\right) }^{\prime }$ ：

\begin{matrix} (10.3.9) & G_{X} (z) = [H (z)] G_{Y} (z) {[H (z^{- 1})]}^{'} . \end{matrix}

10.4 The Spectrum for Vector Processes

Definition of the Population Spectrum for Vector Processes

${\mathbf{y}}_{t}$ $\left( {n \times 1}\right)$ $E\left( {\mathbf{y}}_{t}\right) = \boldsymbol{\mu }$ $k$ 个自协方差矩阵为

\begin{matrix} (10.4.1) & Γ_{k} = E [(y_{t} - μ) {(y_{t - k} - μ)}^{'}] . \end{matrix}

${\left\{ {\boldsymbol{\Gamma }}_{k}\right\} }_{k = - \infty }^{\infty }$ $z$ $y$ 的自协方差生成函数为

\begin{matrix} (10.4.2) & G_{Y} (z) = \sum_{k = - \infty}^{\infty} Γ_{k} z^{k} . \end{matrix}

${\mathbf{G}}_{\mathbf{Y}}\left( z\right)$ $\left( {n \times n}\right)$ $z$ ${2\pi }$ $z = {e}^{-{i\omega }}$ $\omega$ $i = \sqrt{-1}$ $\mathbf{y}$ 的总体谱（population spectrum）：

\begin{matrix} (10.4.3) & s_{Y} (ω) = {(2 π)}^{- 1} G_{Y} (e^{- i ω}) = {(2 π)}^{- 1} \sum_{k = - \infty}^{\infty} Γ_{k} e^{- i ω k} . \end{matrix}

$\left( {n \times n}\right)$ $\omega$ 关联起来。

${\mathbf{s}}_{\mathbf{Y}}\left( \omega \right)$ ${e}^{i\omega k}$ $\omega$ $- \pi$ $\pi$ $\mathbf{y}$ $k$ 个自协方差矩阵的对应元素：

\begin{matrix} (10.4.4) & \int_{- π}^{π} s_{Y} (ω) e^{i ω k} d ω = Γ_{k} . (利 用 E u l e r 公 式 证 明 即 可) \end{matrix}

${\left\{ {\boldsymbol{\Gamma }}_{k}\right\} }_{k = - \infty }^{\infty }$ ${\mathbf{s}}_{\mathbf{Y}}\left( \omega \right)$ 表示的函数包含相同的信息。

$k = 0$ 时，式 [10.4.4] 意味着

\begin{matrix} (10.4.5) & \int_{- π}^{π} s_{Y} (ω) d ω = Γ_{0} . \end{matrix}

$\mathbf{y}$ 的无条件方差-协方差矩阵。

${\boldsymbol{\Gamma }}_{k}$ $j$ $E\left( {{y}_{jt} - {\mu }_{j}}\right) \left( {{y}_{j, t - k} - {\mu }_{j}}\right)$ ${y}_{jt}$ $k$ ${s}_{Y}\left( \omega \right)$ $j$ ${y}_{jt}$ ${s}_{Y}\left( \omega \right)$ $\omega$ ${s}_{Y}\left( \omega \right)$ ${s}_{Y}\left( \omega \right)$ 的非对角元素将是复数。

$n = 2$ 变量的情形，记为

\begin{matrix} y_{t} = [\begin{array}{l} X_{t} \\ Y_{t} \end{array}] . \end{matrix}

$k$ 个自协方差矩阵为

\begin{aligned} (10.4.6) & Γ_{k} & = E [\begin{array}{ll} (X_{t} - μ_{X}) (X_{t - k} - μ_{X}) & (X_{t} - μ_{X}) (Y_{t - k} - μ_{Y}) \\ (Y_{t} - μ_{Y}) (X_{t - k} - μ_{X}) & (Y_{t} - μ_{Y}) (Y_{t - k} - μ_{Y}) \end{array}] \\ \equiv [\begin{array}{ll} γ_{X X}^{(k)} & γ_{X Y}^{(k)} \\ γ_{Y X}^{(k)} & γ_{Y Y}^{(k)} \end{array}] . \end{aligned}

${\boldsymbol{\Gamma }}_{k}^{\prime } = {\boldsymbol{\Gamma }}_{-k}$ 。因此，

\begin{aligned} (10.4.7) & γ_{X X}^{(k)} & = γ_{X X}^{(- k)} \\ (10.4.8) & γ_{Y Y}^{(k)} & = γ_{Y Y}^{(- k)} \\ (10.4.9) & γ_{X Y}^{(k)} & = γ_{Y X}^{(- k)} . \end{aligned}

$n = 2$ 的情形，总体谱 [10.4.3] 为

\begin{aligned} (10.4.10) & s_{Y} (ω) & = \frac{1}{2 π} [\begin{array}{c} \sum_{k = - \infty}^{\infty} γ_{X X}^{(k)} e^{- i ω k} & \sum_{k = - \infty}^{\infty} γ_{X Y}^{(k)} e^{- i ω k} \\ \sum_{k = - \infty}^{\infty} γ_{Y X}^{(k)} e^{- i ω k} & \sum_{k = - \infty}^{\infty} γ_{Y Y}^{(k)} e^{- i ω k} \end{array}] \\ = \frac{1}{2 π} [\begin{array}{c} \sum_{k = - \infty}^{\infty} γ_{X X}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} & \sum_{k = - \infty}^{\infty} γ_{X Y}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} \\ \sum_{k = - \infty}^{\infty} γ_{Y X}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} & \sum_{k = - \infty}^{\infty} γ_{Y Y}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} \end{array}] . \end{aligned}

$\sin \left( {-{\omega k}}\right) = - \sin \left( {\omega k}\right)$ $\sin \left( 0\right) = 0$ 的事实，虚部从对角项中消失：

\begin{matrix} (10.4.11) & \begin{matrix} s_{Y} (ω) = \frac{1}{2 π} [\begin{matrix} \sum_{k = - \infty}^{\infty} γ_{X X}^{(k)} \cos (ω k) & \sum_{k = - \infty}^{\infty} γ_{X Y}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} \\ \sum_{k = - \infty}^{\infty} γ_{Y X}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} & \sum_{k = - \infty}^{\infty} γ_{Y Y}^{(k)} \cos (ω k) \end{matrix}] . \end{matrix} \end{matrix}

${\gamma }_{XY}^{\left( k\right) } \neq {\gamma }_{XY}^{\left( -k\right) }$ ，非对角（off-diagonal）元素通常是复数。

The Cross Spectrum, Cospectrum, and Quadrature Spectrum

$X$ $Y$ 的总体交叉谱（population cross spectrum）：

\begin{matrix} (10.4.12) & s_{Y X} (ω) = {(2 π)}^{- 1} \sum_{k = - \infty}^{\infty} γ_{Y X}^{(k)} {\cos (ω k) - i \cdot \sin (ω k)} . \end{matrix}

交叉谱可以写成其实部和虚部的形式：

\begin{matrix} (10.4.13) & s_{Y X} (ω) = c_{Y X} (ω) + i \cdot q_{Y X} (ω) . \end{matrix}

$X$ $Y$ 之间的协谱（cospectrum）：

\begin{matrix} (10.4.14) & c_{Y X} (ω) = {(2 π)}^{- 1} \sum_{k = - \infty}^{\infty} γ_{Y X}^{(k)} \cos (ω k) . \end{matrix}

$\cos \left( {-{\omega k}}\right) = \cos \left( {\omega k}\right)$ 的事实验证

\begin{matrix} (10.4.15) & c_{Y X} (ω) = c_{X Y} (ω) . \end{matrix}

$X$ $Y$ 的正交谱（quadrature spectrum）：

\begin{matrix} (10.4.16) & q_{Y X} (ω) = - {(2 π)}^{- 1} \sum_{k = - \infty}^{\infty} γ_{Y X}^{(k)} \sin (ω k) . \end{matrix}

$\sin \left( {-{\omega k}}\right) = - \sin \left( {\omega k}\right)$ $Y$ $X$ $X$ $Y$ 的正交谱的负值：

q_{Y X} (ω) = - q_{X Y} (ω) .

${s}_{Y}\left( \omega \right)$ ${s}_{Y}\left( \omega \right)$ $j$ $m$ $m$ $j$ 列元素的复共轭。

${c}_{YX}\left( \omega \right)$ ${q}_{YX}\left( \omega \right)$ $\omega$ 的实值周期函数：

\begin{aligned} c_{Y X} (ω + 2 π j) & = c_{Y X} (ω) for j = \pm 1, \pm 2, \dots \\ q_{Y X} (ω + 2 π j) & = q_{Y X} (ω) for j = \pm 1, \pm 2, \dots . \end{aligned}

进一步，由式 [10.4.14] 可得

c_{Y X} (- ω) = c_{Y X} (ω),

而式 [10.4.16] 意味着

\begin{matrix} (10.4.17) & q_{Y X} (- ω) = - q_{Y X} (ω) . \end{matrix}

$\omega$ $\pi$ 之间的值完全确定了它们。

$X$ $Y$ 之间的无条件协方差：

\int_{- π}^{π} s_{Y X} (ω) d ω = E (Y_{t} - μ_{Y}) (X_{t} - μ_{X}) .

从式 [10.4.17] 观察到，正交谱的积分为零：

\int_{- π}^{π} q_{Y X} (ω) d ω = 0.

$X$ $Y$ $X$ $Y$ 之间的协谱下的面积计算：

\begin{matrix} (10.4.18) & \int_{- π}^{π} c_{Y X} (ω) d ω = E (Y_{t} - μ_{Y}) (X_{t} - μ_{X}) . \end{matrix}

$\omega$ $X$ $Y$ $X$ $Y$ $\omega$ ${c}_{YX}\left( \omega \right)$ 在某些频率上可能为正，在其他频率上可能为负。

综上，通过总体谱我们可以计算自协方差，而通过交叉谱（进一步可以说是协谱）我们可以计算协方差。

The Sample Multivariate Periodogram

${y}_{1},{y}_{2},\ldots ,{y}_{T}$ ${x}_{1},{x}_{2},\ldots ,{x}_{T}$ $T$ $T$ ${y}_{t}$ 的值可以表示为

\begin{matrix} (10.4.19) & y_{t} = \bar{y} + \sum_{j = 1}^{M} {{\hat{α}}_{j} \cdot \cos [ω_{j} (t - 1)] + {\hat{δ}}_{j} \cdot \sin [ω_{j} (t - 1)]}, \end{matrix}

$\bar{y}$ $y$ $M = \left( {T - 1}\right) /2$ ${\omega }_{j} = {2\pi j}/T$ ，且

\begin{aligned} (10.4.20) & {\hat{α}}_{j} & = (2 / T) \sum_{t = 1}^{T} y_{t} \cdot \cos [ω_{j} (t - 1)] \\ (10.4.21) & {\hat{δ}}_{j} & = (2 / T) \sum_{t = 1}^{T} y_{t} \cdot \sin [ω_{j} (t - 1)] . \end{aligned}

$x$ 的类似表示为

\begin{matrix} (10.4.22) & x_{t} = \bar{x} + \sum_{j = 1}^{M} {{\hat{a}}_{j} \cdot \cos [ω_{j} (t - 1)] + {\hat{d}}_{j} \cdot \sin [ω_{j} (t - 1)]} \end{matrix}

其中

\begin{aligned} (10.4.23) & {\hat{a}}_{j} & = (2 / T) \sum_{t = 1}^{T} x_{t} \cdot \cos [ω_{j} (t - 1)] \\ (10.4.24) & {\hat{d}}_{j} & = (2 / T) \sum_{t = 1}^{T} x_{t} \cdot \sin [ω_{j} (t - 1)] . \end{aligned}

回忆式 [6.2.11]，式 [10.4.19] 中的周期回归变量都具有样本均值零且相互正交，同时

\begin{matrix} (10.4.25) & \sum_{t = 1}^{T} \cos^{2} [ω_{j} (t - 1)] = \sum_{t = 1}^{T} \sin^{2} [ω_{j} (t - 1)] = T / 2. \end{matrix}

$x$ $y$ 之间的样本协方差：

\begin{matrix} (10.4.26) & T^{- 1} \sum_{t = 1}^{T} (y_{t} - \bar{y}) (x_{t} - \bar{x}) . \end{matrix}

将式 [10.4.19] 和式 [10.4.22] 代入式 [10.4.26] 并利用周期回归变量的相互正交性，可得

\begin{aligned} (10.4.27) & T^{- 1} \sum_{t = 1}^{T} (y_{t} - \bar{y}) (x_{t} - \bar{x}) & = T^{- 1} \sum_{t = 1}^{T} {\sum_{j = 1}^{M} {{\hat{α}}_{j} \cdot \cos [ω_{j} (t - 1)] + {\hat{δ}}_{j} \cdot \sin [ω_{j} (t - 1)]} \\ \times \sum_{j = 1}^{M} {{\hat{a}}_{j} \cdot \cos [ω_{j} (t - 1)] + {\hat{d}}_{j} \cdot \sin [ω_{j} (t - 1)]]} \\ = T^{- 1} \sum_{t = 1}^{T} {\sum_{j = 1}^{M} {{\hat{α}}_{j} {\hat{a}}_{j} \cdot \cos^{2} [ω_{j} (t - 1)] + {\hat{δ}}_{j} {\hat{d}}_{j} \cdot \sin^{2} [ω_{j} (t - 1)]}} \\ = (1 / 2) \sum_{j = 1}^{M} ({\hat{α}}_{j} {\hat{a}}_{j} + {\hat{δ}}_{j} {\hat{d}}_{j}) . \end{aligned}

上式等号的成立利用了离散傅里叶正交基的正交性，详细可以参考补充材料。

$x$ $y$ ${\omega }_{j}$ 的周期的共同依赖的部分由下式给出：

\begin{matrix} (10.4.28) & (1 / 2) ({\hat{α}}_{j} {\hat{a}}_{j} + {\hat{δ}}_{j} {\hat{d}}_{j}) . \end{matrix}

这个量可以通过与建立命题 6.2 的结果 (c) 时使用的类似计算与协谱的样本类似量联系起来。回忆由于

\sum_{t = 1}^{T} \cos [ω_{j} (t - 1)] = 0,

${\widehat{\alpha }}_{j}$ 可以替代地表示为

{\hat{α}}_{j} = (2 / T) \sum_{t = 1}^{T} (y_{t} - \bar{y}) \cdot \cos [ω_{j} (t - 1)] .

因此，

\begin{aligned} ({\hat{a}}_{j} + i \cdot {\hat{d}}_{j}) ({\hat{α}}_{j} - i \cdot {\hat{δ}}_{j}) \\ = (\frac{4}{T^{2}}) {\sum_{t = 1}^{T} (x_{t} - \bar{x}) \cdot \cos [ω_{j} (t - 1)] + i \cdot \sum_{t = 1}^{T} (x_{t} - \bar{x}) \cdot \sin [ω_{j} (t - 1)]} \\ \times {\sum_{τ = 1}^{T} (y_{τ} - \bar{y}) \cdot \cos [ω_{j} (τ - 1)] - i \cdot \sum_{τ = 1}^{T} (y_{τ} - \bar{y}) \cdot \sin [ω_{j} (τ - 1)]} \\ = (\frac{4}{T^{2}}) {\sum_{t = 1}^{T} (x_{t} - \bar{x}) \cdot \exp [i \cdot ω_{j} (t - 1)]} {\sum_{τ = 1}^{T} (y_{τ} - \bar{y}) \cdot \exp [- i \cdot ω_{j} (τ - 1)]} \\ = (\frac{4}{T^{2}}) {\sum_{t = 1}^{T} (x_{t} - \bar{x}) (y_{t} - \bar{y}) + \sum_{t = 1}^{T - 1} (x_{t} - \bar{x}) (y_{t + 1} - \bar{y}) \cdot \exp [- i ω_{j}] \\ + \sum_{t = 2}^{T} (x_{t} - \bar{x}) (y_{t - 1} - \bar{y}) \cdot \exp [i ω_{j}] + \sum_{t = 1}^{T - 2} (x_{t} - \bar{x}) (y_{t + 2} - \bar{y}) \cdot \exp [- 2 i ω_{j}] \\ + \sum_{t = 3}^{T} (x_{t} - \bar{x}) (y_{t - 2} - \bar{y}) \cdot \exp [2 i ω_{j}] + \dots \\ + (x_{1} - \bar{x}) (y_{T} - \bar{y}) \cdot \exp [- (T - 1) i ω_{j}] \\ + (x_{T} - \bar{x}) (y_{1} - \bar{y}) \cdot \exp [(T - 1) i ω_{j}]} \\ = (\frac{4}{T}) {{\hat{γ}}_{y x}^{(0)} + {\hat{γ}}_{y x}^{(1)} \cdot \exp [- i ω_{j}] + {\hat{γ}}_{y x}^{(- 1)} \cdot \exp [i ω_{j}] \\ + {\hat{γ}}_{y x}^{(2)} \cdot \exp [- 2 i ω_{j}] + {\hat{γ}}_{y x}^{(- 2)} \cdot \exp [2 i ω_{j}] + \dots \\ (10.4.29) & + {\hat{γ}}_{y x}^{(T - 1)} \cdot \exp [- (T - 1) i ω_{j}] + {\hat{γ}}_{y x}^{(- T + 1)} \cdot \exp [(T - 1) i ω_{j}]}, \end{aligned}

${\widehat{\gamma }}_{yx}^{\left( k\right) }$ $y$ $x$ $k$ 期之前的值之间的样本协方差：

\begin{matrix} (10.4.30) & \begin{matrix} {\hat{γ}}_{y x}^{(k)} = {\begin{cases} (1 / T) \sum_{t = 1}^{T - k} (x_{t} - \bar{x}) (y_{t + k} - \bar{y}) & for k = 0, 1, 2, \dots, T - 1 \\ (1 / T) \sum_{t = - k + 1}^{T} (x_{t} - \bar{x}) (y_{t + k} - \bar{y}) & for k = - 1, - 2, \dots, - T + 1. \end{cases} \end{matrix} \end{matrix}

结果 [10.4.29] 意味着

\begin{aligned} (10.4.31) & \frac{1}{2} ({\hat{a}}_{j} + i \cdot {\hat{d}}_{j}) ({\hat{α}}_{j} - i \cdot {\hat{δ}}_{j}) & = (2 / T) \sum_{k = - T + 1}^{T - 1} {\hat{γ}}_{y x}^{(k)} \cdot \exp [- k i ω_{j}] \\ = (4 π / T) \cdot {\hat{s}}_{y x} (ω_{j}), \end{aligned}

${s}_{yx}\left( {\omega }_{j}\right)$ ${\omega }_{j}$ $x$ $y$ 的样本交叉周期图，或样本多元周期图的左下元素：

\begin{matrix} {\hat{s}}_{y} (ω) = {(2 π)}^{- 1} [\begin{matrix} \sum_{k = - T + 1}^{T - 1} {\hat{γ}}_{x x}^{(k)} e^{- i ω k} & \sum_{k = - T + 1}^{T - 1} {\hat{γ}}_{x y}^{(k)} e^{- i ω k} \\ \sum_{k = - T + 1}^{T - 1} {\hat{γ}}_{y x}^{(k)} e^{- i ω k} & \sum_{k = - T + 1}^{T - 1} {\hat{γ}}_{y y}^{(k)} e^{- i ω k} \end{matrix}] = [\begin{matrix} {\hat{s}}_{x x} (ω) & {\hat{s}}_{x y} (ω) \\ {\hat{s}}_{y x} (ω) & {\hat{s}}_{y y} (ω) \end{matrix}] . \end{matrix}

${\omega }_{j}$ $x$ $y$ 的样本交叉周期图可以表示为

\begin{aligned} (10.4.32) & {\hat{S}}_{y x} (ω_{j}) & = [T / (8 π)] \cdot ({\hat{a}}_{j} + i \cdot {\hat{d}}_{j}) ({\hat{α}}_{j} - i \cdot {\hat{δ}}_{j}) \\ = [T / (8 π)] \cdot ({\hat{a}}_{j} {\hat{α}}_{j} + {\hat{d}}_{j} {\hat{δ}}_{j}) + i \cdot [T / (8 π)] \cdot ({\hat{d}}_{j} {\hat{α}}_{j} - {\hat{a}}_{j} {\hat{δ}}_{j}) . \end{aligned}

实部是协谱的样本类似量，而虚部是正交谱的样本类似量：

\begin{matrix} (10.4.33) & {\hat{s}}_{y x} (ω_{j}) = {\hat{c}}_{y x} (ω_{j}) + i \cdot {\hat{q}}_{y x} (ω_{j}), \end{matrix}

其中

\begin{aligned} (10.4.34) & {\hat{c}}_{y x} (ω_{j}) & = [T / (8 π)] \cdot ({\hat{a}}_{j} {\hat{α}}_{j} + {\hat{d}}_{j} {\hat{δ}}_{j}) \\ (10.4.35) & {\hat{q}}_{y x} (ω_{j}) & = [T / (8 π)] \cdot ({\hat{d}}_{j} {\hat{α}}_{j} - {\hat{a}}_{j} {\hat{δ}}_{j}) . \end{aligned}

${\omega }_{j}$ $y$ $x$ ${\omega }_{j}$ $Y$ $X$ $\omega$ 的周期的部分。

那么我们应该如何解释正交谱呢？

${x}_{t}^{ * }$ ：

\begin{aligned} (10.4.36) & x_{t}^{*} & = \bar{x} + \sum_{j = 1}^{M} {{\hat{a}}_{j} \cdot \cos [ω_{j} (t - 1) + (π / 2)] \\ + {\hat{d}}_{j} \cdot \sin [ω_{j} (t - 1) + (π / 2)]} . \end{aligned}

${x}_{t}^{ * }$ ${x}_{t}$ $t = 1$ ${x}_{t}$ 的情况那样刚刚开始。

$\sin \left\lbrack {\theta + \left( {\pi /2}\right) }\right\rbrack = \cos \left( \theta \right)$ $\cos \left\lbrack {\theta + \left( {\pi /2}\right) }\right\rbrack = - \sin \left( \theta \right)$ ${x}_{t}^{ * }$ 可以替代地描述为

x_{t}^{*} = \bar{x} + \sum_{j = 1}^{M} {{\hat{d}}_{j} \cdot \cos [ω_{j} (t - 1)] - {\hat{a}}_{j} \cdot \sin [ω_{j} (t - 1)]} .

${y}_{t}$ ${x}_{t}^{ * }$ 之间的样本协方差为

T^{- 1} \sum_{t = 1}^{T} (y_{t} - \bar{y}) (x_{t}^{*} - \bar{x}) = (1 / 2) \sum_{j = 1}^{M} ({\hat{α}}_{j} {\hat{d}}_{j} - {\hat{δ}}_{j} {\hat{a}}_{j}) .

${\omega }_{j}$ $x$ $y$ ${x}^{ * }$ $y$ ${\omega }_{j}$ ${\omega }_{j}$ $x$ $y$ ${s}_{xx}\left( \omega \right)$ ${\widehat{s}}_{yy}\left( \omega \right)$ $x$ $y$ 更早响应经济衰退。正交谱可以提供寻找这种不同相位周期的证据。

Coherence, Phase, and Gain

$X$ $Y$ $X$ $Y$ $\omega$ 的周期影响的度量。该度量结合了协谱和正交谱的推断，定义为

h_{Y X} (ω) = \frac{{[c_{Y X} (ω)]}^{2} + {[q_{Y X} (ω)]}^{2}}{s_{Y Y} (ω) s_{X X} (ω)},

${s}_{YY}\left( \omega \right)$ ${s}_{XX}\left( \omega \right)$ ${s}_{YY}\left( \omega \right)$ ${s}_{XX}\left( \omega \right)$ $X$ $Y$ $\omega$ $0 \leq {h}_{YX}\left( \omega \right) \leq 1$ ${h}_{YX}\left( \omega \right)$ $Y$ $X$ $\omega$ 的重要共同周期。

$X$ $Y$ 的总体交叉谱写为

\begin{matrix} (10.4.37) & s_{Y X} (ω) = c_{Y X} (ω) + i \cdot q_{Y X} (ω) = R (ω) \cdot \exp [i \cdot θ (ω)], \end{matrix}

其中

\begin{aligned} (10.4.38) & R (ω) & = {{[c_{Y X} (ω)]}^{2} + {[q_{Y X} (ω)]}^{2}}^{1 / 2} \\ (10.4.39) & \frac{\sin [θ (ω)]}{\cos [θ (ω)]} & = \frac{q_{Y X} (ω)}{c_{Y X} (ω)}, \end{aligned}

$\theta \left( \omega \right)$ $R\left( \omega \right)$ $\theta \left( \omega \right)$ 被称为相位（phase）。

$MA$ $AR$ Processes

${\mathbf{y}}_{t}$ ${MA}\left( \infty \right)$ 过程：

y_{t} = μ + Ψ (L) ε_{t},

其中

\begin{matrix} E (ε_{t} ε_{τ}^{'}) = {\begin{cases} Ω & for t = τ \\ 0 & otherwise \end{cases} . \end{matrix}

${\mathbf{y}}_{t}$ 的总体谱可以计算为

\begin{matrix} (10.4.40) & s_{Y} (ω) = {(2 π)}^{- 1} [Ψ (e^{- i ω})] Ω {[Ψ (e^{i ω})]}^{'} . \end{matrix}

${VAR}\left( p\right)$ 的总体谱为

\begin{aligned} (10.4.41) & s_{Y} (ω) & = {(2 π)}^{- 1} {I_{n} - Φ_{1} e^{- i ω} - Φ_{2} e^{- 2 i ω} - \dots - Φ_{p} e^{- p i ω}}^{- 1} Ω \\ \times {I_{n} - Φ_{1}^{'} e^{i ω} - Φ_{2}^{'} e^{2 i ω} - \dots - Φ_{p}^{'} e^{p i ω}}^{- 1} . \end{aligned}

Estimating the Population Spectrum

${\mathbf{y}}_{1},{\mathbf{y}}_{2},\ldots ,{\mathbf{y}}_{T}$ $p$ 阶向量自回归描述，估计总体谱的一个好方法是通过 OLS 估计向量自回归 [10.1.4] 的参数，然后将这些参数估计值代入式 [10.4.41]。

${\omega }_{j} = {2\pi j}/T$ $x$ $y$ ${\widehat{\alpha }}_{j},{\widehat{\delta }}_{j},{\widehat{a}}_{j}$ ${\widehat{d}}_{j}$ ${\omega }_{j}$ $X$ $Y$ 之间的总体协谱的一个合理估计为

{\hat{c}}_{Y X} (ω_{j}) = \sum_{m = - h}^{h} {\frac{h + 1 - | m |}{{(h + 1)}^{2}}} {\hat{c}}_{y x} (ω_{j + m}),

${\widehat{c}}_{yx}\left( {\omega }_{j + m}\right)$ ${\omega }_{j + m} = {2\pi }\left( {j + m}\right) /T$ $h$ ${\omega }_{j}$ 处的协谱时要使用多少个不同频率。

Chapter 6 的公式 [6.3.5] 阐述了这个核函数的意义。

${\kappa }_{k}^{ * }$ ${\widehat{\boldsymbol{\Gamma }}}_{k}$ 。这样的估计将采用以下形式：

{\hat{s}}_{Y} (ω) = {(2 π)}^{- 1} {{\hat{Γ}}_{0} + \sum_{k = 1}^{T - 1} κ_{k}^{*} [{\hat{Γ}}_{k} e^{- i ω k} + {\hat{Γ}}_{k}^{'} e^{i ω k}]},

其中

\begin{aligned} {\hat{Γ}}_{k} & = T^{- 1} \sum_{t = k + 1}^{T} (y_{t} - \overset{―}{y}) {(y_{t - k} - \overset{―}{y})}^{'} \\ \overset{―}{y} & = T^{- 1} \sum_{t = 1}^{T} y_{t} . \end{aligned}

例如，多元谱的修正 Bartlett 估计为

\begin{matrix} (10.4.42) & {\hat{s}}_{Y} (ω) = {(2 π)}^{- 1} {{\hat{Γ}}_{0} + \sum_{k = 1}^{q} [1 - \frac{k}{q + 1}] [{\hat{Γ}}_{k} e^{- i ω k} + {\hat{Γ}}_{k}^{'} e^{i ω k}]} . \end{matrix}

Filters

${\mathbf{x}}_{t}$ $r$ $\left( {r \times r}\right)$ ${s}_{\mathbf{x}}\left( \omega \right)$ ${\left\{ {\mathbf{H}}_{k}\right\} }_{k = - \infty }^{\infty }$ $\left( {n \times r}\right)$ ${\mathbf{y}}_{t}$ $n$ 维向量过程：

y_{t} = H (L) x_{t} = \sum_{k = - \infty}^{\infty} H_{k} x_{t - k} .

$\mathbf{y}$ ${\mathbf{s}}_{\mathbf{Y}}\left( \omega \right)$ $\mathbf{x}$ 的总体谱的关系为

\begin{matrix} (10.4.43) & s_{Y} (ω) = [H (e^{- i ω})] s_{X} (ω) {[H (e^{i ω})]}^{'} . \end{matrix}

${X}_{t}$ ${s}_{X}\left( \omega \right)$ ${u}_{t}$ ${s}_{U}\left( \omega \right)$ ${X}_{t}$ ${u}_{\tau }$ $t$ $\tau$ ${\mathbf{x}}_{t} = {\left( {X}_{t},{u}_{t}\right) }^{\prime }$ 的总体谱由下式给出：

\begin{matrix} s_{X} (ω) = [\begin{matrix} s_{X X} (ω) & 0 \\ 0 & s_{U U} (ω) \end{matrix}] . \end{matrix}

${Y}_{t}$ ：

\begin{matrix} (10.4.44) & Y_{t} = \sum_{k = - \infty}^{\infty} h_{k} X_{t - k} + u_{t} \equiv h (L) X_{t} + u_{t}, \end{matrix}

${\left\{ {h}_{k}\right\} }_{k = - \infty }^{\infty }$ ${\mathbf{y}}_{t} \equiv {\left( {X}_{t},{Y}_{t}\right) }^{\prime }$ ${\mathbf{x}}_{t}$ 获得：

y_{t} = H (L) x_{t},

其中

\begin{matrix} H (L) = [\begin{matrix} 1 & 0 \\ h (L) & 1 \end{matrix}] . \end{matrix}

$y$ 的谱由下式给出：

\begin{aligned} (10.4.45) & s_{Y} (ω) & = [\begin{array}{c} 1 & 0 \\ h (e^{- i ω}) & 1 \end{array}] [\begin{array}{c} s_{X X} (ω) & 0 \\ 0 & s_{U U} (ω) \end{array}] [\begin{array}{c} 1 & h (e^{i ω}) \\ 0 & 1 \end{array}] \\ = [\begin{array}{c} s_{X X} (ω) & s_{X X} (ω) h (e^{i ω}) \\ h (e^{- i ω}) s_{X X} (ω) & h (e^{- i ω}) s_{X X} (ω) h (e^{i ω}) + s_{U U} (ω) \end{array}], \end{aligned}

其中

\begin{matrix} (10.4.46) & h (e^{- i ω}) = \sum_{k = - \infty}^{\infty} h_{k} e^{- i ω k} . \end{matrix}

${Y}_{t}$ ${X}_{t}$ $X$ $Y$ $X$ 的谱来计算。

${\mathbf{y}}_{t} = {\left( {X}_{t},{Y}_{t}\right) }^{\prime }$ ，其具有绝对可和的自协方差矩阵，总体谱由下式给出：

\begin{matrix} (10.4.47) & \begin{matrix} s_{Y} (ω) = [\begin{array}{ll} s_{X X} (ω) & s_{X Y} (ω) \\ s_{Y X} (ω) & s_{Y Y} (ω) \end{array}] . \end{matrix} \end{matrix}

${Y}_{t}$ ${\left\{ {X}_{t - k}\right\} }_{k = - \infty }^{\infty }$ ${u}_{t}$ ${\left\{ {h}_{k}\right\} }_{k = - \infty }^{\infty }$ $\omega$ 的函数来概括。比较式 [10.4.47] 和式 [10.4.45] 的左下元素，该函数必须满足

h (e^{- i ω}) s_{X X} (ω) = s_{Y X} (ω) .

$h\left( {e}^{-{i\omega }}\right)$ 可以从下式计算：

\begin{matrix} (10.4.48) & h (e^{- i ω}) = \frac{s_{Y X} (ω)}{s_{X X} (ω)} ⟶ transfer function \end{matrix}

${s}_{XX}\left( \omega \right)$ ${s}_{XX}\left( \omega \right) = 0$ $h\left( {e}^{-{i\omega }}\right) = 0$ $X$ $Y$ $X$ $X$ $Y$ 的传递函数（transfer function）。

式 [10.4.4] 的基本原理可以进一步用于揭示各个传递函数系数：

h_{k} = {(2 π)}^{- 1} \int_{- π}^{π} h (e^{- i ω}) e^{i ω k} d ω .

式 [10.4.46] 将传递函数展开为傅里叶级数：
$h\left( {e}^{-{i\omega }}\right) = \sum_{k = - \infty }^{\infty }{h}_{k}{e}^{-{i\omega k}}.$
${e}^{i\omega j}$ $\left[ - \pi ,\pi \right]$ 上积分：
$\int_{-\pi }^{\pi }h\left( {e}^{-{i\omega }}\right) {e}^{i\omega j}{d\omega } = \int_{-\pi }^{\pi }\sum_{k = - \infty }^{\infty }{h}_{k}{e}^{-{i\omega k}}{e}^{i\omega j}{d\omega } = \sum_{k = - \infty }^{\infty }{h}_{k}\int_{-\pi }^{\pi }{e}^{i\omega \left( {j - k}\right) }{d\omega }.$
利用复指数函数的正交性：
$\int_{-\pi }^{\pi }{e}^{i\omega \left( {j - k}\right) }{d\omega } = \begin{cases} 2\pi & \text{当 }j = k \\ 0 & \text{当 }j \neq k \end{cases},$
$\int_{-\pi }^{\pi }h\left( {e}^{-{i\omega }}\right) {e}^{i\omega j}{d\omega } = 2\pi \cdot {h}_{j},$ ${h}_{j} = {\left( 2\pi \right) }^{-1}\int_{-\pi }^{\pi }h\left( {e}^{-{i\omega }}\right) {e}^{i\omega j}{d\omega }.$ $j$ $k$ 即得上述公式。

${\left( {X}_{t},{Y}_{t}\right) }^{\prime }$ ${Y}_{t}$ ${\left\{ {X}_{t - k}\right\} }_{k = - \infty }^{\infty }$ ${X}_{t - k}$ 的系数可以从下式计算：

\begin{matrix} (10.4.49) & h_{k} = {(2 π)}^{- 1} \int_{- π}^{π} \frac{s_{Y X} (ω)}{s_{X X} (ω)} e^{i ω k} d ω . \end{matrix}

10.5 The Sample Mean of a Vector Process

Variance of the Sample Mean

$T$ $\left\{ {{\mathbf{y}}_{1},{\mathbf{y}}_{2},\ldots ,{\mathbf{y}}_{T}}\right\}$ $n$ 维协方差平稳过程：

\begin{aligned} (10.5.1) & E (y_{t}) & = μ \\ (10.5.2) & E [(y_{t} - μ) {(y_{t - j} - μ)}^{'}] & = Γ_{j} . \end{aligned}

考虑样本均值的性质：

\begin{matrix} (10.5.3) & {\overset{―}{y}}_{T} = (1 / T) \sum_{t = 1}^{T} y_{t} . \end{matrix}

$E\left( {\overline{\mathbf{y}}}_{T}\right) = \boldsymbol{\mu }$ ，且

\begin{aligned} E [({\overset{―}{y}}_{T} - μ) {({\overset{―}{y}}_{T} - μ)}^{'}] & = (1 / T^{2}) E {(y_{1} - μ) [{(y_{1} - μ)}^{'} + {(y_{2} - μ)}^{'} + \dots + {(y_{T} - μ)}^{'}] \\ + (y_{2} - μ) [{(y_{1} - μ)}^{'} + {(y_{2} - μ)}^{'} + \dots + {(y_{T} - μ)}^{'}] \\ + (y_{3} - μ) [{(y_{1} - μ)}^{'} + {(y_{2} - μ)}^{'} + \dots + {(y_{T} - μ)}^{'}] \\ (10.5.4) & + \dots + (y_{T} - μ) [{(y_{1} - μ)}^{'} + {(y_{2} - μ)}^{'} + \dots + {(y_{T} - μ)}^{'}]} \\ = (1 / T^{2}) {[Γ_{0} + Γ_{- 1} + \dots + Γ_{- (T - 1)}] \\ + [Γ_{1} + Γ_{0} + Γ_{- 1} + \dots + Γ_{- (T - 2)}] \\ + [Γ_{2} + Γ_{1} + Γ_{0} + Γ_{- 1} + \dots + Γ_{- (T - 3)}] \\ + \dots + [Γ_{T - 1} + Γ_{T - 2} + Γ_{T - 3} + \dots + Γ_{0}]} \\ = (1 / T^{2}) {T Γ_{0} + (T - 1) Γ_{1} + (T - 2) Γ_{2} + \dots + Γ_{T - 1} \\ + (T - 1) Γ_{- 1} + (T - 2) Γ_{- 2} + \dots + Γ_{- (T - 1)}} . \end{aligned}

因此，

\begin{aligned} (10.5.5) & T \cdot E [({\overset{―}{y}}_{T} - μ) {({\overset{―}{y}}_{T} - μ)}^{'}] & = Γ_{0} + [(T - 1) / T] Γ_{1} + [(T - 2) / T] Γ_{2} + \dots \\ + [1 / T] Γ_{T - 1} + [(T - 1) / T] Γ_{- 1} + [(T - 2) / T] Γ_{- 2} \\ + \dots + [1 / T] Γ_{- (T - 1)} . \end{aligned}

$\left| k\right|$ ${\boldsymbol{\Gamma }}_{k}$ $T \rightarrow \infty$ 趋于 1，对于协方差平稳过程，更高的自协方差趋于零。因此，我们有命题 7.5 的以下推广。

Important

命题 10.5 ： ${\mathbf{y}}_{t}$ 为协方差平稳过程，矩由式 [10.5.1] 和式 [10.5.2] 给出，且具有绝对可和的自协方差。则样本均值 [10.5.3] 满足

${\overline{\mathbf{y}}}_{T}\overset{p}{ \rightarrow }\boldsymbol{\mu }$

$\lim_{T \rightarrow \infty }\left\{ {T \cdot E\left\lbrack {\left( {{\overline{\mathbf{y}}}_{T} - \boldsymbol{\mu }}\right) {\left( {\overline{\mathbf{y}}}_{T} - \boldsymbol{\mu }\right) }^{\prime }}\right\rbrack }\right\} = \sum_{v = - \infty }^{\infty }{\boldsymbol{\Gamma }}_{v}$ .

证明： $\left( {n \times n}\right)$ 矩阵：

\begin{matrix} (10.5.6) & \sum_{v = - \infty}^{\infty} Γ_{v} - T \cdot E [({\overset{―}{y}}_{T} - μ) {({\overset{―}{y}}_{T} - μ)}^{'}] = \sum_{| v | \geq T} Γ_{v} + \sum_{v = - (T - 1)}^{T - 1} (| v | / T) Γ_{v}, \end{matrix}

${\gamma }_{ij}^{\left( v\right) }$ ${\boldsymbol{\Gamma }}_{v}$ $i$ $j$ $i$ $j$ 列元素可以写成

\sum_{| v | \geq T} γ_{i j}^{(v)} + \sum_{v = - (T - 1)}^{T - 1} (| v | / T) γ_{i j}^{(v)} .

${\left\{ {\boldsymbol{\Gamma }}_{v}\right\} }_{v = - \infty }^{\infty }$ $\varepsilon > 0$ $q$ 使得

\sum_{| v | > q} | γ_{i j}^{(v)} | < ε / 2.

因此，

| \sum_{| v | \geq T} γ_{i j}^{(v)} + \sum_{v = - (T - 1)}^{T - 1} (| v | / T) γ_{i j}^{(v)} | < ε / 2 + \sum_{v = - q}^{q} (| v | / T) | γ_{i j}^{(v)} | .

$T$ $\varepsilon$ $i$ $E{\left( {\bar{y}}_{i, T} - {\mu }_{i}\right) }^{2} \rightarrow 0$ ${\bar{y}}_{i, T}\overset{p}{ \rightarrow }{\mu }_{i}$ 。证毕。

Chapter 10 —— Covariance-Stationary Vector Processes

10.1 Introduction to Vector Autoregressions

Rewriting a VAR(p)VAR(p) as a VAR(1)VAR(1)

Conditions for Stationarity

Vector MA(∞)\mathrm{MA}\left( \infty \right) Representation

Assumptions Implicit in a VAR

10.2 Autocovariances and Convergence Results for Vector Processes

The jjth Autocovariance Matrix

Vector MA(q)MA(q) Process

Vector MA(∞)MA(\infty) Process

Multivariate Filters

Vector Autoregression

10.3 The Autocovariance-Generating Function for Vector Processes

Definition of Autocovariance-Generating Function for Vector Processes

Autocovariance-Generating Function for a Vector Moving Average Process

Autocovariance-Generating Function for a Vector Autoregression

Transformations of Vector Processes

10.4 The Spectrum for Vector Processes

Definition of the Population Spectrum for Vector Processes

The Cross Spectrum, Cospectrum, and Quadrature Spectrum

The Sample Multivariate Periodogram

Coherence, Phase, and Gain

The Population Spectrum for Vector MAMA and ARAR Processes

Estimating the Population Spectrum

Filters

10.5 The Sample Mean of a Vector Process

Variance of the Sample Mean

$VAR(p)$ $VAR(1)$

$\mathrm{MA}\left( \infty \right)$ Representation

$j$ th Autocovariance Matrix

$MA(q)$ Process

$MA(\infty)$ Process

$MA$ $AR$ Processes